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### Transcript of Solving equations involving exponents and logarithms

• Solving equations involving exponents and logarithms

• Lets review some terms.

When we write log5 125

5 is called the base125 is called the argument

• Logarithmic form of 52 = 25 is

log525 = 2

• For all the laws a, M and N > 0a 1r is any real

• Remember ln and logln is a short cut for logelog means log10

• Log laws

• If your variable is in an exponent or in the argument of a logarithmFind the pattern your equation resembles

• If your variable is in an exponent or in the argument of a logarithmFind the pattern Fit your equation to match the patternSwitch to the equivalent formSolve the resultCheck (be sure you remember the domain of a log)

• log(2x) = 3

• It fits log(2x) = 3

• Switch log(2x) = 3103=2xDid you remember that log(2x) means log10(2x)?

• Divide by 2 log(2x) = 3103=2x500 = x

• ln(x+3) = ln(-7x)

• ln(x+3) = ln(-7x) It fits

• ln(x+3) = ln(-7x)Switch

• ln(x+3) = ln(-7x)x + 3 = -7x

Switch

• ln(x+3) = ln(-7x)x + 3 = -7xx = -

Solve the result(and check)

• ln(x) + ln(3) = ln(12)

• ln(x) + ln(3) = ln(12)x + 3 = 12

• ln(x) + ln(3) = ln(12)x + 3 = 12 Oh NO!!! Thats wrong!

• You need to use log lawsln(x) + ln(3) = ln(12) ln(3x) = ln (12)

• Switchln(x) + ln(3) = ln(12)ln(3x) = ln (12)3x = 12

• ln(x) + ln(3) = ln(12)ln(3x) = ln (12)3x = 12x = 4Solve the result

• log3(x+2) + 4 = 9

• It will fit log3(x+2) + 4 = 9

• Subtract 4 to make it fit log3(x+2) + 4 = 9log3(x+2) = 5

• Switch log3(x+2) + 4 = 9log3(x+2) = 5

• Switch log3(x+2) + 4 = 9log3(x+2) = 535 = x + 2

• Solve the resultlog3(x+2) + 4 = 9log3(x+2) = 535 = x + 2 x = 241

• 5(10x) = 19.45

• Divide by 5 to fit5(10x) = 19.4510x = 3.91

• Switch 5(10x) = 19.4510x = 3.91

• Switch 5(10x) = 19.4510x = 3.91log(3.91) = x

• Exact log(3.91) Approx 0.5925(10x) = 19.4510x = 3.91log(3.91) = x 0.592

• 2 log3(x) = 8

• It will fit 2 log3(x) = 8

• Divide by 2 to fit 2 log3(x) = 8log3(x) = 4

• Switch 2 log3(x) = 8log3(x) = 4

• Switch 2 log3(x) = 8log3(x) = 434=x

• Then Simplify2 log3(x) = 8log3(x) = 434=x x = 81

• log2(x-1) + log2(x-1) = 3

• Need to use a log lawlog2(x-1) + log2(x-1) = 3

• log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 3

• Switchlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 3

• Switchlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1)

• and finishlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

• But -3 does not check!log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

• Exclude -3 (it would cause you to have a negative argument)log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

• Theres more than one way to do this

• Can you find why each step is valid?

• rules of exponents multiply both sides by 2

• Heres another way to solve the same equation.

• exclude 2nd resultSquare both sidesSimplify

• 52x - 5x 12 = 0

• Factor it. Think of y2 - y-12=052x - 5x 12 = 0(5x 4)(5x + 3) = 0

• Set each factor = 052x - 5x 12 = 0(5x 4)(5x + 3) = 05x 4 = 0 or 5x + 3 = 0

• Solve first factors equationSolve 5x 4 = 0 5x = 4log54 = x

• Solve other factors equationSolve 5x + 3 = 0 5x = -3log5(-3) = x

• Oops, we cannot have a negative argumentSolve 5x + 3 = 0 5x = -3log5(-3) = x

• Exclude this solution.Only the other factors solution worksSolve 5x + 3 = 0 5x = -3log5(-3) = x

• 4x+2 = 5x

• If M = N then ln M = ln N

4x+2 = 5x

• If M = N then ln M = ln N4x+2 = 5x ln(4x+2) = ln(5x )

• 4x+2 = 5x ln(4x+2) = ln(5x)

• 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )

• Distribute 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)

• Get x terms on one side 4x+2 = 5x ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4)

• Factor out x 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4)

• Divide by numerical coefficient 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4)